Relations and Functions (Class 11) · JEE Preparation Mathematics
Let f: R → R be defined by f(x) = (x − 1)² + 2. Find the range of f and state whether f is one-one.
Practice question from Relations and Functions (Class 11) (JEE Preparation Mathematics), with the correct answer and a step-by-step solution.
medium · Subjective · Q-80FF25
Let f: R → R be defined by f(x) = (x − 1)² + 2. Find the range of f and state whether f is one-one.
Answer: Since (x − 1)² ≥ 0 for every real x, f(x) ≥ 2, and f(1) = 2; hence the range of f is [2, ∞). The function is not one-one because f(0) = f(2) = 3.
Solution
The square term is always non-negative, so the least value of is 2. Equal values at two distinct inputs, such as , show that the function is not one-one.
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